将 responseObject base64 编码字符串转换为 UIImage
Converting responseObject base64 encoded string to UIImage
我正在尝试将带有 PHP 的图像(在服务器端)发送到我的 iOS 应用程序,以便我可以在 UIImageView 中显示它。
我的服务器端代码是:
<?php
header("Content-Type: image/jpeg"); //if your data is format jpeg
$username = $_POST['username'];
$count = $_POST['count'];
$base64string = base64_encode(file_get_contents("images/".$username."/".$count."/".$username."file".$count.".jpeg"));
echo $base64string;
?>
我在 iOS 应用程序中收到带有此代码的图像:
NSString * uploadURL = @"http://192.168.1.4/getimage.php";
NSLog(@"uploadImageURL: %@", uploadURL);
NSString *queryStringss = [NSString stringWithFormat:@"%@", uploadURL];
queryStringss = [queryStringss stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
manager.responseSerializer=[AFJSONResponseSerializer serializerWithReadingOptions:NSJSONReadingAllowFragments];
manager.responseSerializer.acceptableContentTypes = [manager.responseSerializer.acceptableContentTypes setByAddingObject:@"text/html"];
manager.responseSerializer.acceptableContentTypes = [manager.responseSerializer.acceptableContentTypes setByAddingObject:@"text/plain"];
manager.responseSerializer.acceptableContentTypes = [manager.responseSerializer.acceptableContentTypes setByAddingObject:@"image/jpeg"];
NSUserDefaults *userdefaults = [NSUserDefaults standardUserDefaults];
NSString *usernameEncoded = [marker.title stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSDictionary *params = @{@"username": usernameEncoded, @"count": [object valueForKey:@"count"]};
[manager POST:queryStringss parameters:params success:^(AFHTTPRequestOperation * _Nonnull operation, id _Nonnull responseObject) {
NSLog(@"Success: %@ ***** %@", operation.responseString, responseObject);
NSData *decodedData = [[NSData alloc] initWithBase64EncodedString:responseObject options:0];
image.image = [UIImage imageWithData:decodedData scale:300/2448];
[self.view addSubview:image];
}
failure:^(AFHTTPRequestOperation *operation, NSError *error) {
NSLog(@"Error: %@ ***** %@", operation.responseString, error);
}];
当我 运行 代码时 - 它遇到了错误块并出现错误,该错误显示为我发送的 base64 编码 "string"(图像):
2015-12-03 01:19:15.655 sneek[6261:1952572] Error: /9j/4AAQSkZJRgABAQAASABIAAD/4QBYRXhpZgAATU0AKgAAAAgAAgESAAMAAAABAAYAAIdpAAQAAAABAAAAJgAAAAAAA6ABAAMAAAABAAEAAKACAAQAAAABAAAMwKADAAQAAAABAAAJkAAAAAD/7QA4UGhvdG9zaG9wIDMuMAA4QklNBAQAAAAAAAA4QklNBCUAAAAAABDUHYzZjwCyBOmACZjs+EJ+/8AAEQgJkAzAAwEiAAIRAQMRAf/EAB8AAAEFAQEBAQEBAAAAAAAAAAABAgMEBQYHCAkKC//EALUQAAIBAwMCBAMFBQQEAAABfQECAwAEEQUSITFBBhNRYQcicRQygZGhCCNCscEVUtHwJDNicoIJChYXGBkaJSYnKCkqNDU2Nzg5OkNERUZHSElKU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6g4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tr
... (very long) ...
Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 0." UserInfo={NSDebugDescription=Invalid value around character 0.}
我做错了什么?
您指定了 AFJSONResponseSerializer
,尽管它不是 JSON。当然,您已经覆盖了 acceptableContentTypes
,但这并不能阻止它尝试在响应中解析 JSON。
我会用 AFHTTPResponseSerializer
然后丢掉 acceptableContentTypes
。
顺便说一句,我不会将 image/jpeg
用于 base64 编码的响应,因为它是文本,而不是 jpeg。如果你要 return 原始 base64 字符串,你可以使用 application/text
或类似的东西。
或者,更好的是,将您的 PHP 实际更改为 return JSON(因为这样可以更轻松地解析响应)并保留 AFJSONResponseSerializer
(但是修复 header 后丢失 acceptableContentTypes
),然后您可以从 response[@"image"]
获取 base64 字符串。
<?php
header("Content-Type: application/json");
$username = $_POST['username'];
$count = $_POST['count'];
$base64string = base64_encode(file_get_contents("images/".$username."/".$count."/".$username."file".$count.".jpeg"));
echo json_encode(array("image" => $base64string));
?>
或者,使用 AFImageResponseSerializer
并将 PHP 更改为 return 图像:
<?php
header("Content-Type: image/jpeg"); //if your data is format jpeg
$username = $_POST['username'];
$count = $_POST['count'];
$contents = file_get_contents("images/".$username."/".$count."/".$username."file".$count.".jpeg"));
echo $contents;
?>
我正在尝试将带有 PHP 的图像(在服务器端)发送到我的 iOS 应用程序,以便我可以在 UIImageView 中显示它。
我的服务器端代码是:
<?php
header("Content-Type: image/jpeg"); //if your data is format jpeg
$username = $_POST['username'];
$count = $_POST['count'];
$base64string = base64_encode(file_get_contents("images/".$username."/".$count."/".$username."file".$count.".jpeg"));
echo $base64string;
?>
我在 iOS 应用程序中收到带有此代码的图像:
NSString * uploadURL = @"http://192.168.1.4/getimage.php";
NSLog(@"uploadImageURL: %@", uploadURL);
NSString *queryStringss = [NSString stringWithFormat:@"%@", uploadURL];
queryStringss = [queryStringss stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
manager.responseSerializer=[AFJSONResponseSerializer serializerWithReadingOptions:NSJSONReadingAllowFragments];
manager.responseSerializer.acceptableContentTypes = [manager.responseSerializer.acceptableContentTypes setByAddingObject:@"text/html"];
manager.responseSerializer.acceptableContentTypes = [manager.responseSerializer.acceptableContentTypes setByAddingObject:@"text/plain"];
manager.responseSerializer.acceptableContentTypes = [manager.responseSerializer.acceptableContentTypes setByAddingObject:@"image/jpeg"];
NSUserDefaults *userdefaults = [NSUserDefaults standardUserDefaults];
NSString *usernameEncoded = [marker.title stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSDictionary *params = @{@"username": usernameEncoded, @"count": [object valueForKey:@"count"]};
[manager POST:queryStringss parameters:params success:^(AFHTTPRequestOperation * _Nonnull operation, id _Nonnull responseObject) {
NSLog(@"Success: %@ ***** %@", operation.responseString, responseObject);
NSData *decodedData = [[NSData alloc] initWithBase64EncodedString:responseObject options:0];
image.image = [UIImage imageWithData:decodedData scale:300/2448];
[self.view addSubview:image];
}
failure:^(AFHTTPRequestOperation *operation, NSError *error) {
NSLog(@"Error: %@ ***** %@", operation.responseString, error);
}];
当我 运行 代码时 - 它遇到了错误块并出现错误,该错误显示为我发送的 base64 编码 "string"(图像):
2015-12-03 01:19:15.655 sneek[6261:1952572] Error: /9j/4AAQSkZJRgABAQAASABIAAD/4QBYRXhpZgAATU0AKgAAAAgAAgESAAMAAAABAAYAAIdpAAQAAAABAAAAJgAAAAAAA6ABAAMAAAABAAEAAKACAAQAAAABAAAMwKADAAQAAAABAAAJkAAAAAD/7QA4UGhvdG9zaG9wIDMuMAA4QklNBAQAAAAAAAA4QklNBCUAAAAAABDUHYzZjwCyBOmACZjs+EJ+/8AAEQgJkAzAAwEiAAIRAQMRAf/EAB8AAAEFAQEBAQEBAAAAAAAAAAABAgMEBQYHCAkKC//EALUQAAIBAwMCBAMFBQQEAAABfQECAwAEEQUSITFBBhNRYQcicRQygZGhCCNCscEVUtHwJDNicoIJChYXGBkaJSYnKCkqNDU2Nzg5OkNERUZHSElKU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6g4SFhoeIiYqSk5SVlpeYmZqio6Slpqeoqaqys7S1tr
... (very long) ...
Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 0." UserInfo={NSDebugDescription=Invalid value around character 0.}
我做错了什么?
您指定了 AFJSONResponseSerializer
,尽管它不是 JSON。当然,您已经覆盖了 acceptableContentTypes
,但这并不能阻止它尝试在响应中解析 JSON。
我会用 AFHTTPResponseSerializer
然后丢掉 acceptableContentTypes
。
顺便说一句,我不会将 image/jpeg
用于 base64 编码的响应,因为它是文本,而不是 jpeg。如果你要 return 原始 base64 字符串,你可以使用 application/text
或类似的东西。
或者,更好的是,将您的 PHP 实际更改为 return JSON(因为这样可以更轻松地解析响应)并保留 AFJSONResponseSerializer
(但是修复 header 后丢失 acceptableContentTypes
),然后您可以从 response[@"image"]
获取 base64 字符串。
<?php
header("Content-Type: application/json");
$username = $_POST['username'];
$count = $_POST['count'];
$base64string = base64_encode(file_get_contents("images/".$username."/".$count."/".$username."file".$count.".jpeg"));
echo json_encode(array("image" => $base64string));
?>
或者,使用 AFImageResponseSerializer
并将 PHP 更改为 return 图像:
<?php
header("Content-Type: image/jpeg"); //if your data is format jpeg
$username = $_POST['username'];
$count = $_POST['count'];
$contents = file_get_contents("images/".$username."/".$count."/".$username."file".$count.".jpeg"));
echo $contents;
?>